有限数学示例

$9 x - 6 Y + 2 Z = - 3$ , $5 x - y + 2 = 5$ , $y - 4 z = - 2$

解题步骤 1

Move variables to the left and constant terms to the right.

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解题步骤 1.1

移动 $9 x$ 。

$- 6 Y + 2 Z + 9 x = - 3$

$5 x - y + 2 = 5$

$y - 4 z = - 2$

解题步骤 1.2

将所有不包含变量的项移到等式右边。

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解题步骤 1.2.1

从等式两边同时减去 $2$ 。

$- 6 Y + 2 Z + 9 x = - 3$

$5 x - y = 5 - 2$

$y - 4 z = - 2$

解题步骤 1.2.2

从 $5$ 中减去 $2$ 。

$- 6 Y + 2 Z + 9 x = - 3$

$5 x - y = 3$

$y - 4 z = - 2$

$- 6 Y + 2 Z + 9 x = - 3$

$5 x - y = 3$

$y - 4 z = - 2$

$- 6 Y + 2 Z + 9 x = - 3$

$5 x - y = 3$

$y - 4 z = - 2$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} - 6 & 2 & 9 & 0 & 0 & - 3 \\ 0 & 0 & 5 & - 1 & 0 & 3 \\ 0 & 0 & 0 & 1 & - 4 & - 2 \end{array}]$

解题步骤 3

求行简化阶梯形矩阵。

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解题步骤 3.1

Multiply each element of $R_{1}$ by $- \frac{1}{6}$ to make the entry at $1, 1$ a $1$ .

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解题步骤 3.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{6}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{6} \cdot - 6 & - \frac{1}{6} \cdot 2 & - \frac{1}{6} \cdot 9 & - \frac{1}{6} \cdot 0 & - \frac{1}{6} \cdot 0 & - \frac{1}{6} \cdot - 3 \\ 0 & 0 & 5 & - 1 & 0 & 3 \\ 0 & 0 & 0 & 1 & - 4 & - 2 \end{array}]$

解题步骤 3.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - \frac{1}{3} & - \frac{3}{2} & 0 & 0 & \frac{1}{2} \\ 0 & 0 & 5 & - 1 & 0 & 3 \\ 0 & 0 & 0 & 1 & - 4 & - 2 \end{array}]$

解题步骤 3.2

Multiply each element of $R_{2}$ by $\frac{1}{5}$ to make the entry at $2, 3$ a $1$ .

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解题步骤 3.2.1

Multiply each element of $R_{2}$ by $\frac{1}{5}$ to make the entry at $2, 3$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{3} & - \frac{3}{2} & 0 & 0 & \frac{1}{2} \\ \frac{0}{5} & \frac{0}{5} & \frac{5}{5} & \frac{- 1}{5} & \frac{0}{5} & \frac{3}{5} \\ 0 & 0 & 0 & 1 & - 4 & - 2 \end{array}]$

解题步骤 3.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{1}{3} & - \frac{3}{2} & 0 & 0 & \frac{1}{2} \\ 0 & 0 & 1 & - \frac{1}{5} & 0 & \frac{3}{5} \\ 0 & 0 & 0 & 1 & - 4 & - 2 \end{array}]$

解题步骤 3.3

Perform the row operation $R_{2} = R_{2} + \frac{1}{5} R_{3}$ to make the entry at $2, 4$ a $0$ .

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解题步骤 3.3.1

Perform the row operation $R_{2} = R_{2} + \frac{1}{5} R_{3}$ to make the entry at $2, 4$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{3} & - \frac{3}{2} & 0 & 0 & \frac{1}{2} \\ 0 + \frac{1}{5} \cdot 0 & 0 + \frac{1}{5} \cdot 0 & 1 + \frac{1}{5} \cdot 0 & - \frac{1}{5} + \frac{1}{5} \cdot 1 & 0 + \frac{1}{5} \cdot - 4 & \frac{3}{5} + \frac{1}{5} \cdot - 2 \\ 0 & 0 & 0 & 1 & - 4 & - 2 \end{array}]$

解题步骤 3.3.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{1}{3} & - \frac{3}{2} & 0 & 0 & \frac{1}{2} \\ 0 & 0 & 1 & 0 & - \frac{4}{5} & \frac{1}{5} \\ 0 & 0 & 0 & 1 & - 4 & - 2 \end{array}]$

解题步骤 3.4

Perform the row operation $R_{1} = R_{1} + \frac{3}{2} R_{2}$ to make the entry at $1, 3$ a $0$ .

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解题步骤 3.4.1

Perform the row operation $R_{1} = R_{1} + \frac{3}{2} R_{2}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 + \frac{3}{2} \cdot 0 & - \frac{1}{3} + \frac{3}{2} \cdot 0 & - \frac{3}{2} + \frac{3}{2} \cdot 1 & 0 + \frac{3}{2} \cdot 0 & 0 + \frac{3}{2} (- \frac{4}{5}) & \frac{1}{2} + \frac{3}{2} \cdot \frac{1}{5} \\ 0 & 0 & 1 & 0 & - \frac{4}{5} & \frac{1}{5} \\ 0 & 0 & 0 & 1 & - 4 & - 2 \end{array}]$

解题步骤 3.4.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - \frac{1}{3} & 0 & 0 & - \frac{6}{5} & \frac{4}{5} \\ 0 & 0 & 1 & 0 & - \frac{4}{5} & \frac{1}{5} \\ 0 & 0 & 0 & 1 & - 4 & - 2 \end{array}]$

解题步骤 4

Use the result matrix to declare the final solution to the system of equations.

$Y - \frac{1}{3} Z - \frac{6}{5} z = \frac{4}{5}$

$x - \frac{4}{5} z = \frac{1}{5}$

$y - 4 z = - 2$

解题步骤 5

The solution is the set of ordered pairs that make the system true.

$(\frac{4}{5} + \frac{Z}{3} + \frac{6 z}{5}, Z, \frac{1}{5} + \frac{4 z}{5}, - 2 + 4 z, z)$

有限数学 示例

有限数学示例